What is the largest angle in the triangle with sides 5, 12 and 13?

What is the largest angle in the triangle with sides 5, 12 and 13?

First of all we should check if the triangle is rectangular.

\(5^2+12^2=13^2\)

\(25+144=169\)

169=169

The triangle is rectangular, then the biggest angle will be 90^{o}

Bases of the trapezium are 7 cm and 15 cm, and a height is 8 cm. Find the area of a trapezium.

\(S=\frac{7+15}{2}\times8=22\times4=88 sm^2\)

The hypotenuse AB of a right triangle ABC is 8 cm, the angle A is 60^{o}. Find the legs and an acute angle.

"a" is the leg opposite the 60° angle. "b" is the other leg.

sin60=\(\frac{\sqrt{3}}{2}=\frac{a}{8}\); \(a=4\sqrt{3}\) cm

cos60=0,5=b\8; b=4 cm.

180°-(90°+60°)=30° is the last angle.

The area of a square with a diagonal of 6 cm, is

"a" is the side of the square. Two sides and the diagonal is a right triangle. Then \(a^2+a^2=6^2; 2a^2=36; a^2=18\)

The area of the square is \(S=a^2=18 \ cm^2\)

The length of the rectangle is increased by 25%. What percent is necessary to reduce the width to its area has not changed?

Solve the task

"a" is the length, "b" is the width.

S=ab at the begining.

The new area is S=1,25a(xb)=ab.

\(x=\frac{1}{1,25}=0,8\)

The width was reduced by 20%.

\(sin\frac{3\pi}{4}\times{cos\frac{\pi}{4}}+cos\frac{3\pi}{4}\times{sin\frac{\pi}{4}}\)

Compute

The area of the triangle is 40 cm^{2}. Height is 5 times lower then the side on which it is lowered. Then the height is

Area: S=0,5ah=40

where "a" is for the side and "h" is the height. (h>0; a>0)

Then a=5h.

S=0,5h(5h)=40

\(5h^2=80\)

\(h^2=16\); h=4

Find the perimeter of a rhombus, if its side is 2 m

Rhombus has equal sides, then its perimeter is P=4*2=8 m

The vector \(\overrightarrow{c}\)(2m;-3) is the sum of the vectors \(\overrightarrow{a}\)(7m;5) and \(\overrightarrow{b}\)(-18;2n) when

Find m and n

2m=7m-18 => m=3,6

-3=5+2n => n=-4

The middle line of the trapezoid is equal to 7 cm. One of the bases is more than another for 4 cm. Find the bases of the trapezoid.

"a" and "b" are the bases of the trapezium.

b=a+4

(a+a+4)/2=7

a+2=7

a=5

b=9

\(ctg\beta-\frac{cos\beta-1}{sin\beta}\)

Simplify

One of the angles of the parallelogram constitutes a quarter of the other, then the smaller angle is equal to

"a" and "b" are the angles of the parallelogram.

a=0,25b; b=4a

a+b=180

a+4a=180; 5a=180; a=36°

If the sum of the two acute angles of a triangle is equal to 90, then a triangle is

\(\angle{A}+\angle{B}=90\)°

\(\angle{A}+\angle{B}+\angle{C}=180\)°

\(\angle{C}=180-90=90\)°

The triangle is rectangular.

Obtuse angle of the parallelogram is 130°. Find the outside angle at an acute angle of the parallelogram.

The outer angle will be equal to the obtuse angle, 130°.

Find the area of the triangle if its height is 10 cm, its base is 5 cm.

\(S=\frac{1}{2}ah=0,5\times10\times5=25 cm^2\)

One of the sides of the rectangle is 5 cm larger than the other. Find the sides of the rectangle, if its area is 14 cm^{2}.

a - one side.

a+5 - another side.

a(a+5)=14

\(a^2+5a-14=0\)

a=-7 (not fit) or

a=2; a+5=7

The area of a square is equal to 8 cm. Then its diagonal is

a - the side of the square. d - the diagonal.

a^{2}=8

d^{2}=2a^{2}

\(d=\sqrt{2a^2}=\sqrt{2\times8}=\sqrt{16}=4\)

For how many degrees 30% of the right angle is less than 20% of the straight angle?

90°*0,3=27°

180°*0,2=36°

36°-27°=9°

The sides of the rectangle are treated as 4 to 9, and its area is 144 cm^{2}. Then a large side is

a and b - the sides.

a/b=4/9; a=4b/9

ab=144

4b^{2}/9=144

b^{2}=324

b=18; a=8

How many sides does a regular polygon have if its every angle is 150°?

\(\frac{180(n-2)}{n}=150\)

180n-360=150n

30n=360

n=12

One of the angles of a parallelogram is equal to 40°. Find the angle that not equal to it.

The sum of twa unequal angles is 180°.

180°-40°=140°

The perimeter of the rhombus is equal to 24 cm, and the area is 24 cm^{2}. Then the height of the rhombus is

P=4a=24; a=6

S=ah=24; h=24/a=24/6=4 cm

The equation of the line passing through the point M (7, -11) and parallel to the line y = -2x + 5 is

y=kx+b - the equation of the line.

Parallel lines have equal k.

Then k=-2

y=-2x+b. It passes through the point M, then:

-11=-2*7+b

b=3

y=-2x+3

The height of the rhombus is equal to 4 cm, and its area is 24 cm^{2}. Then the perimeter of the rhombus is

S=ah=24 cm^{2}

h=4 cm; a=24:h=24:4=6 cm

P=4a=4*6=24 cm

The perimeter of the square is 24 cm. Find the length of the sides of a rectangle having the same perimeter, knowing that one side of the rectangle is 2 times larger than the other.

a and 2a - the sides of the rectangle.

2(a+2a)=24

6a=24

a=4 cm

2a=8 cm

The segment on the map has a length of 3.2 cm, and on the terrain - 1,6 km. Then the segment corresponding to the distance on the terrain of 2,8 km, on the same map will have a length of

x - the distance on the map

\(\frac{3,2}{1,6}=\frac{x}{2,8}\)

\(x = \frac{3,2\times2,8}{1,6}\)

x=5,6

Squares of two similar pentagons are in the ration of 4:9. Find the ratio of their perimeters.

Squares of two similar pentagons are in the ration of 4:9. Find the ratio of their perimeters.

Find the perimeter of a rectangle, if the perpendiculars drawn from the point of intersection of the diagonals of the sides of a rectangle are equal to 2 and 4 dm.

Find the perimeter of a rectangle, if the perpendiculars drawn from the point of intersection of the diagonals of the sides of a rectangle are equal to 2 dm and 4 dm.

These perpendiculars will be equal to the half of corresponding sides of the rectangle. Then sides of the rectangle will be 4 and 8 dm.

The perimeter is: P=2(4+8)=24 dm

The area of the rectangle is 42 cm^{2}. If the width of the rectangle is 3 cm, then what is its length?

^{2}. If the width of the rectangle is 3 cm, then what is its length?

x - is the length of the rectangle.

Then

3x=42;

x=14 cm

The radius of the inscribed circle in a square is 2 cm. Find the area of the square.

The radius of the inscribed circle in a square is 2 cm. Find the area of the square.

The side of the square is equal to 2r, where "r" is the radius of the inscribed circle.

Then

S=a^{2}=(2r)^{2}=4^{2}=16 cm^{2}

\(cos\alpha\times{tg}\alpha-sin\alpha\)

Simplify

The length of the rectangle is 40 cm and a width is 60% of its length. Find the width of the rectangle.

40*0,6=24 cm

Compute cos300^{0}

Compute cos300^{0}

\(cos300^0=cos(360^0-60^0)=cos60^0=\frac12=0,5\)

What is the area of a rectangle, if the diagonal of 6 cm forms with the side an angle of 60.

What is the area of a rectangle, if the diagonal of 6 cm forms with the side an angle of 60.

The sides are "a" and "b"

a=6sin60^{0}

b=6cos60^{0}

S=ab=\(6sin60^0\times6cos60^0=36\times\frac12\times\frac{\sqrt3}{2}=9\sqrt3\) cm^{2}

If cosa=0,8; then tga could be

If cosa=0,8; then tga could be

\(sina=\pm\sqrt{1-cos^2a}=\pm\sqrt{1-\frac{16}{25}}=\pm\frac34\)

\(tga=\frac{sina}{cosa}=\frac{\pm\frac35}{\frac45}=\pm\frac34=\pm0,75\)